Question

A mapping to $N{\text{ }}to{\text{ }}N$ is defined as follows
$f:{\text{ }}N{\text{ }} - - - > N$
$f(n){\text{ = (n + 5}}{{\text{)}}^2},n\varepsilon N$
($N$ is a set of natural numbers). Then,
A.$F$ is not one to one
B.$F$ is onto
C.$F$ is both one to one and onto
D.$F$ is one to one but not onto

Answer 1

Hint: - So we have to find the cases of one-one and onto.
To check onto, we check that the inverse of the function and also the codomain exist in that interval.
To check one-one, we have to check that the function does not give the same value for two different points.

Complete step by step solution:
We are given
Now, ${1^{st}}$ we will check for one-one,
$f\left( {n\\_1} \right) = f\left( {n\\_2} \right)$
Let ${n_1},{n_2}\varepsilon N$
Now, assume
${({n_1} - 5)^2} = {({n_2} - 5)^2}$
$= > n\\_1 - 5 = n\\_2 - 5$
$= > n\\_1 = n\\_2$
$\left( {n\\_1} \right){\text{ }} = \left( {n\\_2} \right)$ $\therefore$ $R$ is one-one
Now, we will check form onto:
${(x - 5)^2} = y$
Now let’s check for the onto condition
$x - 5 = {y^{(1/2)}}$
$x = {y^{(1/2)}} + 5$
Therefore we cannot find the values of $y$ all natural numbers.
Therefore this is not onto.
$R$ is not onto.
Hence, f is one-one but not onto
Therefore, the option $\left( D \right)$ is correct.

Note: - While checking one-one and onto, follow the basic rules that rules $i.e.$for one-one different elements should have different values and for onto range should be equal to domain.